How do you factor #5x^4 + x^3 - 22x^2 - 4x + 8#?

1 Answer
May 11, 2015

The result is #5x^4+x^3-22x^2-4x+8=5(x+2)(x-2)(x-((-1+sqrt41)/10))(x-((-1-sqrt41)/10))#.

The procedure is the following:

You have to apply Ruffini's Rule trying the divisors of the independent term (in this case, the divisors of 8) until you find one that makes the rest of the division zero.
I started with +1 and -1 but it did not work, but if you try (-2) you get it:

          ! 5     1      -22    -4    8
      -2!      -10     +18    +8   -8
            5   -9      -4      +4    0

What you have here is that #5x^4+x^3-22x^2-4x+8=(x+2)(5x^3-9x^2-4x+4)#. [By the way, remember that if you have succeeded in applying Ruffini's Rule with a certain number "a" (in this case, with (-2)), you have to write the factor as (x-a) (in this case, (x-(-2)), which is (x+2)].

Now you have got one factor (x+2) and you have to keep going the same process with #5x^3-9x^2-4x+4#.
If you try now with +2 you will get it:

             ! 5     -9      -4      4
          2 !        10       2     -4
                5     +1      -2     0

So, what you have now is that #5x^3-9x^2-4x+4=(x-2)(5x^2+x-2)#.

And summing up what we have done until now:

Now, you have got two factors: (x+2) and (x-2) and you have to decompose #5x^2+x-2#.

In this case, instead of applying Ruffini's Rule, we shall apply the classic resolution formula to the quadratic equation: #5x^2+x-2=0#, which will be: #x= (-1+-sqrt(1^2-4(5)(-2)))/10= ((-1)+-sqrt(41))/10#, and that will give you two solutions:
#x_1=((-1)+sqrt41)/10# and #x_2=((-1)-sqrt41)/10#, which are the two last factors.

So what we have now is that #5x^2+x-2=5(x-(-1+sqrt41)/10)(x-(-1-sqrt41)/10)# [note that the factorization needs to be multiplied by the coefficient of the #x^2#].

So the solution is: #5x^4+x^3-22x^2-4x+8=5(x+2)(x-2)(x-(-1+sqrt41)/10)(x-(-1-sqrt41)/10)#.